SOLUTION: A store sells three types of flash drives: brand A, brand B, and brand C. Of the flash drives it sells, 60% are brand A, 25% are brand B, and 15% are brand C. The store has found t
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-> SOLUTION: A store sells three types of flash drives: brand A, brand B, and brand C. Of the flash drives it sells, 60% are brand A, 25% are brand B, and 15% are brand C. The store has found t
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Question 1161236: A store sells three types of flash drives: brand A, brand B, and brand C. Of the flash drives it sells, 60% are brand A, 25% are brand B, and 15% are brand C. The store has found that 20% of the brand A flash drives, 15% of the brand B flash drives, and 5% of the brand C flash drives are returned as defective. If a flash drive is returned as defective, what is the probability that it is a brand A flash drive? A brand B flash drive? A brand C flash drive? Answer by ikleyn(52781) (Show Source):
It is a conditional probability problem.
The probability that the flash drive is defective is
P(defective) = 0.60*0.2 + 0.25*0.15 + 0.15*0.05 = 0.165.
The probability that the randomly selected flash drive, made by machine A is defective, is
P(A and defective) = 0.60*0.2 = 0.12.
The probability that the randomly selected defective flash drive was made by machine A is
P(A | defective) = P(A and defective)/P(defective) = = = 0.727273. ANSWER to (a)
Similarly, the probability that the randomly selected defective flash drive was made by machine B is
P(B | defective) = P(B and defective)/P(defective) = = 0.227273. ANSWER to (b)
the probability that the randomly selected defective flash drive was made by machine C is
P(C | defective) = P(C and defective)/P(defective) = = 0.045455. ANSWER to (c)
As a check, the sum of the three conditional probabilities is equal to 1 (one).
0.727273 + 0.227273 + 0.045455 = 1.
